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Andreyy89
3 years ago
9

What average power must be supplied to the rope to generate sinusoidal waves that have amplitude 0.200 m and wavelength 0.600 m

if the waves are to travel at 28.0 m/s ?
Physics
1 answer:
pychu [463]3 years ago
6 0

Complete question:

A taut rope has a mass of 0.123 kg and a length of 3.54 m. What average power must be supplied to the rope to generate sinusoidal waves that have amplitude 0.200 m and wavelength 0.600 m if the waves are to travel at 28.0 m/s ?

Answer:

The average power supplied to the rope to generate sinusoidal waves is 1676.159 watts.

Explanation:

Velocity = Frequency  X wavelength

V = Fλ ⇒ F = V/λ

F = 28/0.6 = 46.67 Hz

Angular frequency (ω) = 2πF = 2π (46.67) = 93.34π rad/s

Average power supplied to the rope will be calculated as follows

P_{avg} =\frac{1}{2} \mu \omega^2 A^2 V

where;

ω is the angular frequency

A is the amplitude

V is the velocity

μ is mass per unit length = 0.123/3.54 = 0.0348 kg/m

P_{avg} =\frac{1}{2} ( 0.0348)(93.34 \pi )^2 (0.2)^2 (28) = 1676.159 watts

The average power supplied to the rope to generate sinusoidal waves is 1676.159 watts.

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Light of wavelength 436.1 nm falls on two slits spaced 0.31 mm apart. What is the required distance from the slits to the screen
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Answer:

The correct answer is "4.26 m".

Explanation:

Given:

Wavelength,

\lambda = 436.1 \ nm

or,

  =436.1\times 10^{-9} \ m

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d = 0.31 \ mm

or,

  =0.31\times 10^{-3} \ m

Distance between the 1st and 2nd dark fringes,

(y_2-y_1) = 6\times 10^{-3} \ m

As we know,

⇒ \frac{d}{L} (y_2-y_1) = \lambda

or,

⇒ L=\frac{d(y_2-y_1)}{\lambda}

By substituting the values, we get

       =\frac{0.31\times 6\times 10^{-6}}{436.1\times 10^{-9}}

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       =\frac{1860}{436.1}

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3 years ago
A rope, under a tension of 221 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement
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3 years ago
Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton
pishuonlain [190]

Hello!

Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton force is applied to it.

Data:

Hooke represented mathematically his theory with the equation:

F = K * Δx  

On what:

F (elastic force) = 2 N

K (elastic constant) = 4 N/cm

Δx (deformation or elongation of the elastic medium or distance from a spring) = ?

Solving:


F = K * \Delta{x}

2\:N = 4\:N/cm*\Delta{x}

4\:N/cm*\Delta{x} = 2\:N

\Delta{x} = \dfrac{2\:\diagup\!\!\!\!\!N}{4\:\diagup\!\!\!\!\!N/cm}

simplify by 2

\Delta{x} = \dfrac{2}{4}\frac{\div2}{\div2}

\boxed{\boxed{\Delta{x} = \dfrac{1}{2}\:cm}}\Longleftarrow(distance)\end{array}}\qquad\checkmark

Answer:

B.) 1/2 cm

_______________________

I Hope this helps, greetings ... Dexteright02! =)

7 0
3 years ago
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